TO be saturated you need an amount of solvent that will not dissolve anymore into the solution. So, the best answer from the list is C
2Al (s) + 3Fe(NO3)2 (aq) → 3Fe (s) + 2Al(NO3)3 (aq)
Find the pure amount of iron nitrate solution:
325 g x 87.5% = 284.375 g
Convert 284.375g into mols by dividing by molar mass of the iron nitrate solution
Molar mass of iron nitrate = 55.85 g/mol + 2x14g/mol + 6x16g/mol = 179.85 g/mol
Moles of iron nitrate = 284.375 g / 179.85 g/mol = 1.58
Stoichiometry
3 mols of iron nitrate / 3 mols of iron = 1.58 mols of iron nitrate / x mols of iron
x = 1.58 mols of iron.
Convert that in grams by multiplying by the molar mass of iron
grams of iron = 1.58 mol x 55.85 g/mol = 88.24 g
Answer:
Around 17,000,000 or a little more than that
B.) Decomposition
hope this helps
